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Algebra 8 / 12

Find the minimum value of the expression $f(x, y, z) = \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$ for any positive real numbers $x, y, z$.

(A) 1

(B) 2

(C) 3

(D) Cannot be determined

Correct Answer: (C)

1. Identify the Relevant Theorem: Since $x, y, z$ are positive real numbers and we are looking for the minimum value of a sum of terms where the product is constant, the Arithmetic Mean - Geometric Mean (AM-GM) Inequality is applicable.

2. State the AM-GM Inequality: For any $n$ positive real numbers $a_1, a_2, \dots, a_n$, the relationship is given by $\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot \dots \cdot a_n}$.

3. Apply the Inequality to the Expression: Let $a_1 = \frac{x}{y}$, $a_2 = \frac{y}{z}$, and $a_3 = \frac{z}{x}$. The mean of these three terms is $\frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \geq \sqrt[3]{\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x}}$.

4. Simplify the Product: Notice that the product $\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = 1$. Therefore, the geometric mean is $\sqrt[3]{1} = 1$.

5. Calculate the Minimum Value: Multiplying both sides by 3, we get $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq 3 \times 1$. Thus, the minimum value is 3.

6. Condition for Equality: The minimum is attained when all terms are equal, i.e., $\frac{x}{y} = \frac{y}{z} = \frac{z}{x}$, which occurs when $x = y = z$.

Test Prep Tip: In Algebra, whenever you see a sum of reciprocal-like terms and the variables are restricted to positive real numbers, the AM-GM Inequality is almost always the most efficient path to the solution.